Conservation of energy with a mass and pulley system.

In summary, the problem involves two boxes connected by a rope passing over a frictionless pulley. The mass of Box A is 15kg and the mass of Box B is 12kg. The system is initially at rest with the bottom of Box A at a height of 0.85m above the floor. Using conservation of energy, the speed with which Box A will contact the floor can be determined by equating the initial potential energy to the final kinetic energy.
  • #1
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Homework Statement


Two boxes are attached to opposite ends of a rope passing over a frictionless pulley as shown below. The mass of Box A is 15kg and the mass of box B is 12kg. The system is originally at rest with the bottom of box A at a height of o.85m above the floor. When the system is released, the boxes will move. Use conservation of energy to determine the speed with which Box A will contact the floor.


Homework Equations



Eg=mgΔh
Ek= 1/2 mv^2/2
ƩFy=may

The Attempt at a Solution


I started off by drawing free body diagrams of each mass, one at rest, and one in motion.

For mass A:
at rest,
ƩFy=0
Ft+Eg=0
Ft= Eg
= mgΔh
=(15kg)(9.8m/s^2)(0.85m)
Ft=125N

in motion,
ƩFy= may
Fg(A)-Ft= m(A)ay

For mass B:
at rest,
ƩFy=0
Fn-mg=0
Fn=mg
=(12kg)(9.8m/s^2)
Fn=117.6N

in motion,
ƩFy=may
Ft-Fg(B)= m(B)ay

I'm not sure if my original statement of Ft=Eg is accurate... and from this point on I don't know where to go.
 
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  • #2
Why not find the initial and final energies and equate them?
 
  • #3
As in, Etotal= Eg, Etotal'=Ek ?
 
  • #4
System is originally at rest. Hence initial energy = ...
Final energy = ...
 
  • #5




Great job on drawing the free body diagrams and identifying the forces acting on each mass. Your initial statement of Ft=Eg is correct, as they are equal and opposite forces in this system. From here, you can use the conservation of energy principle to solve for the speed with which Box A will contact the floor. This principle states that the total energy of a closed system remains constant, and in this case, the system is only experiencing gravitational potential energy (Eg) and kinetic energy (Ek). Therefore, we can equate the initial potential energy (Eg) to the final kinetic energy (Ek):

Eg = Ek
mgΔh = 1/2 mv^2

Solving for v, we get:

v = √(2gh)

Plugging in the values given in the problem, we get:

v = √(2*9.8m/s^2*0.85m) = 4.38 m/s

Therefore, the speed with which Box A will contact the floor is 4.38 m/s. This solution is only valid if the rope remains taut throughout the motion, and if the pulley is frictionless. Great job on applying the conservation of energy principle to this mass and pulley system!
 

1. What is the conservation of energy in a mass and pulley system?

The conservation of energy in a mass and pulley system refers to the principle that energy cannot be created or destroyed, but can only be transferred or converted from one form to another.

2. How does the mass and pulley system conserve energy?

In a mass and pulley system, the potential energy of the object at a higher position is converted into kinetic energy as it moves downwards due to the force of gravity. This kinetic energy is then converted back into potential energy as the object moves back up due to the tension in the pulley. Therefore, the system conserves energy by continuously converting between potential and kinetic energy.

3. Is the conservation of energy always applicable in a mass and pulley system?

Yes, the conservation of energy is always applicable in a mass and pulley system as long as there are no external forces acting on the system. This means that energy will be conserved even as the system moves and changes, as long as there are no external factors that can affect the energy transformation.

4. How can the conservation of energy in a mass and pulley system be calculated?

The conservation of energy in a mass and pulley system can be calculated by using the equation: PE + KE = constant, where PE is potential energy, KE is kinetic energy, and the constant represents the total energy in the system. This equation can also be written as mgh + 1/2mv^2 = constant, where m is the mass, g is the acceleration due to gravity, h is the height, and v is the velocity of the object.

5. What are some real-life applications of the conservation of energy in a mass and pulley system?

The conservation of energy in a mass and pulley system has many real-life applications, such as in elevators, cranes, and ski lifts. In these systems, the potential energy of the object is converted into kinetic energy as it moves downwards, and then converted back into potential energy as it moves back up. This allows for efficient and continuous movement without the need for additional energy input.

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